T(n)=T(4n/5)+n

Solution for T(n)=T(4n/5)+n equation:

(T)=(4T/5)+T

We move all terms to the left:

(T)-((4T/5)+T)=0


Domain of the equation: 5)+T)!=0

T!=0/1

T!=0

T∈R


We add all the numbers together, and all the variables

T-((+4T/5)+T)=0

We multiply all the terms by the denominator


T*5)+T)-((+4T=0

We add all the numbers together, and all the variables

4T+T*5)+T)-((=0

Wy multiply elements

5T^2+4T=0

a = 5; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·5·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*5}=\frac{-8}{10}
=-4/5
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*5}=\frac{0}{10}
=0
$